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I'm doing homework for Inorganic Chemistry, and I am a little confused.

The Question is:

What is the electronic configuration of $\ce{Ru^{2+}}$?

If I were to remove from the highest energy level it would be $\ce{[Kr]~5s^1~4d^5}$ leaving $6$ unpaired electrons.

If I were to take into account electron pairing it would be $\ce{[Kr]~4d^6}$ with $4$ unpaired electrons.

What would be the correct approach?

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You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$.

All you need to know know is the order of energies for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the Aufbau principle (and possible pitfalls/exceptions) and another thread on this site.

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  • $\begingroup$ As mentioned above i attempted an answer based on the order of energies (I considered that 4d is higher energy than 5s), here is a link for reference. I've considered the Madelung Rule here but it doesn't seem to apply. $\endgroup$ – John Snow Sep 4 '14 at 16:35
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I don't know if it's kosher to provide the answer like this but it turns out (after i got my homework back) that the correct answer was;

$Ru^{2+}: [Kr]$ $4d^6$.

The reasoning provided that was behind it is "When the d or f orbitals start to fill up, the energy of these orbitals is lowered such that they have lower energy than the corresponding s orbitals."

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There is another simple version of Aufbau principle. n+l rule, where n is principle quantum number and l is azimuthal quantum number. For 4d n+l= 4+2=6 and for 5s n+l=5+0=5, so electron will enter 4d orbital first and then it will fill 5s if 4d is filled or half filled. Note: This principle is for filling the orbital only. But for removing electron it must be removed from the outer shell first, in this case 5s. So, electron will be removed from 5s and then 4d and so on. NB: Order of energy should only be applied for filling the orbital and for removing electron it must be removed from the outer shell and so on.

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